Skip to content
EXERCISE 1.5 · Q199

Q.y=log⁡(log⁡2x)y=\log(\log 2x), show that xy2+y1(1+xy1)=0xy_2+y_1(1+xy_1)=0

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
68% · 199/293 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Given y=log⁡(log⁡2x)y=\log(\log 2x). Let L=log⁡(2x)L=\log(2x), so y=log⁡Ly=\log L, and note dLdx=12x⋅2=1x\dfrac{dL}{dx}=\dfrac{1}{2x}\cdot2=\dfrac1x.

Step 1: y1=1L⋅1x=1xLy_1=\dfrac{1}{L}\cdot\dfrac1x=\dfrac{1}{xL}.

Step 2 — differentiate y1=(xL)−1y_1=(xL)^{-1}: y2=−(xL)−2⋅ddx(xL)=−L+x⋅1x(xL)2=−L+1(xL)2y_2=-(xL)^{-2}\cdot\dfrac{d}{dx}(xL)=-\dfrac{L+x\cdot\frac1x}{(xL)^2}=-\dfrac{L+1}{(xL)^2}.

Step 3 — compute xy1=xxL=1Lxy_1=\dfrac{x}{xL}=\dfrac1L, so 1+xy1=1+1L=L+1L1+xy_1=1+\dfrac1L=\dfrac{L+1}{L}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.